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MathGroup Archive 2007

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Re: drawing

  • To: mathgroup at smc.vnet.net
  • Subject: [mg76732] Re: drawing
  • From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
  • Date: Sat, 26 May 2007 04:42:42 -0400 (EDT)
  • Organization: The Open University, Milton Keynes, UK
  • References: <f36gu8$9tn$1@smc.vnet.net>

dimitris wrote:
> The following produce a bounded region like those encountered
> in any elementary vector analysis book.
> 
> In[41]:=
> Clear["Global`*"]
> 
> In[42]:=
> << "Graphics`Arrow`"
> 
> We want to take a circle and add an arbitrary modulation to the
> radius
> to obtain an irregular shape. However, we need a smooth join at o=0
> and o=2Pi. So we need a partition function. I am going to make a
> function that is a over most of the domain 0 to 2 , but smoothly
> transistions to zero at the end points and has zero slope at the end
> points. The following was just used to calculate arguments for the
> function.
> 
> In[43]:=
> a*o + b /. o -> 2*Pi
> Out[43]=
> b + 2*a*Pi
> 
> In[45]:=
> a*o + b /. o -> 2*Pi - d
> Out[45]=
> b + a*(-d + 2*Pi)
> 
> In[46]:=
> Solve[{b + 2*a*Pi == -(Pi/2), b + a*(-d + 2*Pi) == Pi/2}, {a, b}]
> Out[46]=
> {{a -> -(Pi/d), b -> -((d*Pi - 4*Pi^2)/(2*d))}}
> 
> In[49]:=
> 1/2 + (1/2)*Sin[(-o)*(Pi/d) - (d*Pi - 4*Pi^2)/(2*d)]
> FullSimplify[%]
> 
> Out[49]=
> 1/2 - (1/2)*Sin[(o*Pi)/d + (d*Pi - 4*Pi^2)/(2*d)]
> Out[50]=
> Sin[((o - 2*Pi)*Pi)/(2*d)]^2
> 
> So this gives us a partition function. d gives the width of the
> transistion region at each end of the o domain.
> 
> In[51]:=
> partitionfunction[d_][o_] := Piecewise[{{Sin[(Pi*o)/(2*d)]^2,
> Inequality[0, LessEqual, o, Less, d]},
>     {1, Inequality[d, LessEqual, o, Less, 2*Pi - d]}, {Sin[(Pi*(2*Pi
> -
> o))/(2*d)]^2, 2*Pi - d <= o <= 2*Pi}}]
> 
> Let's use a piece of a Bessel function to modulate the radius.
> 
> In[52]:=
> Plot[BesselJ[5, x], {x, 5, 18}, Frame -> True];
> 
> In[53]:=
> Solve[{(a*o + b /. o -> 0) == 5, (a*o + b /. o -> 2*Pi) == 18}]
> Out[53]=
> {{a -> 13/(2*Pi), b -> 5}}
> 
> In[54]:=
> radius[d_][o_] := 1 + 1.5*partitionfunction[d][o]*BesselJ[5, (13/
> (2*Pi))*o + 5]
> 
> In[55]:=
> Plot[radius[1][o], {o, 0, 2*Pi}, Frame -> True, PlotRange -> All,
> Axes
> -> False];
> 
> Now we can parametrize the curve.
> 
> In[58]:=
> curve[d_][o_] := radius[d][o]*{Cos[o], Sin[o]}
> 
> For d=1 and o=45=B0 we can calculate the tangent line and normal
> line.
> 
> In[59]:=
> tangent[t_] = N[curve[1][45*Degree] + t*Derivative[1][curve[1]]
> [45*Degree]]
> Out[59]=
> {1.057382730502271 - 0.7335911589691905*t, 1.057382730502271 +
> 1=2E3811743020353515*t}
> 
> In[61]:=
> normal[t_] = N[curve[1][45*Degree] + t*Reverse[Derivative[1]
> [curve[1]]
> [45*Degree]]*{1, -1}]
> Out[61]=
> {1.057382730502271 + 1.3811743020353515*t, 1.057382730502271 +
> 0=2E7335911589691905*t}
> 
> In[83]:=
> n = {1.127382730502271, 1.037382730502271};
> 
> Finnally
> 
> In[81]:=
> Block[{$DisplayFunction = Identity}, g = ParametricPlot[curve[1][o1],
> {o1, 0, 2*Pi}, Axes -> False, PlotPoints -> 50,
>      PlotStyle -> Thickness[0.007]]; g1 = g /. Line[x_] ->
> {GrayLevel[0.8], Polygon[x]};
>    g2 = ParametricPlot[tangent[t], {t, -0.2, 0.2}, PlotStyle ->
> Thickness[0.006], PlotPoints -> 50];
>    g3 = Graphics[{Thickness[0.007], Arrow[normal[0], normal[0.3],
> HeadLength -> 0.06, HeadCenter -> 0.7]}];
>    cir = Graphics[{Circle[normal[0], 0.1, {3.3*(Pi/2), 2.15*Pi}]}];
> po
> = Graphics[{PointSize[0.01], Point[n]}];
>    tex1 = Graphics[Text["V", {0.0532359, -0.0138103}]]; tex2 =
> Graphics[Text["S", {0.470751, -1.08655}]];
>    tex3 = Graphics[Text[StyleForm["n", FontSize -> 17, FontFamily ->
> "Times", FontColor -> Black,
> FontWeight -> "Bold"], {1.7, 1.2}]]; ]
> 
> Show[g, g1, g2, g3, tex1, tex2, tex3, cir, po, AspectRatio ->
> Automatic,
>    TextStyle -> {FontSize -> 17, FontFamily -> "Times", FontWeight ->
> "Bold"}];
> 
> Murray Eisenberg has tried this code in Mathematica 6 and after made
> the proper
> modifications (Arrow package is obsolete in the new version for
> example) he got
> a very unpleasant graph.
> 
> See here
> 
> http://smc.vnet.net/mgattach/curve.png
> 
> Any ideas what is going one?
> 
> I would very appreciate if someone could modify the code to work in
> version 6 and put the resulting
> graphic in some link page. The antialiasing makes the quality of the
> graphics objects by far more
> superior and "cleaner" in version 6 but unfortunately I don't have it
> yet!
> 
> Thanks
> Dimitris

Hi Dimitris,

I believe that the following is very close to what you expect. I have 
fixed the code for the arrow using the built-in function Arrow and 
Arrowheads (note that the spelling is correct: the second word, head, is 
not capitalized. I would have expected ArrowHeads).

Note that by changing the order of the graphs in the last Show command, 
in version 6.0, (Show[g3, g, g1, g2, ...] rather than Show[g, g1, g2, 
g3, ...]) you get the normal vector correctly displayed. You do not need 
to do that in version 5.2.

You can see the resulting plot at

http://homepages.nyu.edu/~jmg336/Dimitris%20Graphics.png

(I have sent to you the notebook and a png picture.)

Regards,
Jean-Marc

P.S. Here is the code for version 6.0.

  Clear["Global`*"]
  a*o + b /. o -> 2*Pi
  a*o + b /. o -> 2*Pi - d
  Solve[ { b + 2*a*Pi == - ( Pi/2), b + a* ( -d + 2*Pi) == Pi/2}, { a, b}]
  1/2 + ( 1/2)* Sin[ ( -o)* ( Pi/d) - ( d*Pi - 4* Pi^2)/ ( 2*d)]
  FullSimplify[%]
  partitionfunction[d_][o_] :=
  Piecewise[ { { Sin[ ( Pi*o)/ ( 2*d)]^2, Inequality[ 0, LessEqual, o, 
Less,
  d]}, { 1, Inequality[ d, LessEqual, o, Less, 2*Pi - d]},
  { Sin[ ( Pi* ( 2*Pi - o))/ ( 2*d)]^2, 2*Pi - d <= o <= 2*Pi}}]
  Plot[ BesselJ[ 5, x], { x, 5, 18}, Frame -> True]
  Solve[ { ( a*o + b /. o -> 0) == 5, ( a*o + b /. o -> 2*Pi) == 18}]
  radius[d_][o_] := 1 + 1.5* partitionfunction[d][o]*
  BesselJ[ 5, ( 13/ ( 2*Pi))*o + 5]
  Plot[ radius[1][o], { o, 0, 2*Pi}, Frame -> True, PlotRange -> All,
  Axes -> False]
  curve[d_][o_] := radius[d][o]* { Cos[o], Sin[o]}
  tangent[t_] = N[ curve[1][ 45*Degree] +
  t* Derivative[1][ curve[1]][ 45*Degree]]
  normal[t_] = N[ curve[1][ 45*Degree] +
  t* Reverse[ Derivative[1][ curve[1]][ 45*Degree]]* { 1, -1}]
  n = { 1.127382730502271, 1.037382730502271};
  g = ParametricPlot[ curve[1][o1], { o1, 0, 2*Pi}, Axes -> False,
  PlotPoints -> 50, PlotStyle -> Thickness[0.007]];
  g1 = g /. Line[x_] -> { GrayLevel[0.8], Polygon[x]};
  g2 = ParametricPlot[ tangent[t], { t, -0.2, 0.2},
  PlotStyle -> Thickness[0.004], PlotPoints -> 50];
  g3 = Graphics[ { Thickness[0.007], Arrowheads[
  { { 0.06, 1., Graphics[ Polygon[ { { 0, 0}, { -1, -1/3}, { -1/3, 0},
  { -1, 1/3}}], ImageSize -> { 38., Automatic}]}}],
  Arrow[ { normal[0], normal[0.3]}]}];
  cir = Graphics[ { Circle[ normal[0], 0.1, { 3.3* ( Pi/2), 2.15*Pi}]}];
  po = Graphics[ { PointSize[0.01], Point[n]}];
  tex1 = Graphics[ Text[ "V", { 0.0532359, -0.0138103}]];
  tex2 = Graphics[ Text[ "S", { 0.470751, -1.08655}]];
  tex3 = Graphics[ Text[ StyleForm[ "n", FontSize -> 17,
  FontFamily -> "Times", FontColor -> Black, FontWeight -> "Bold"],
  { 1.7, 1.2}]];
  Show[ g3, g, g1, g2, tex1, tex2, tex3, cir, po,
  AspectRatio -> Automatic, TextStyle -> { FontSize -> 17,
  FontFamily -> "Times", FontWeight -> "Bold"}]


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